The correct answer is (D):
Let the area of ABCD be 100. Side of ABCD = 10 Area of EFGH is 62.5 => Side of EFGH = √62.5
Triangles AEH, BFE, CGF and DHG are congruent by ASA.
Let AE = BF = CG = DH = x; EB = FC = DG = AH = 10 -xx
AE2+ AH2 = EH2
x2+(10-x)2=(√62.5)2
Solving, x = 2.5 or 7.5
Since it’s given that CG is longer than EB, CG = 7.5 and EB = 2.5.
Therefore, EB : CG = 1 : 3